The notion of a knot concordance is a rich subject in low-dimensional topology, see Livingston's survey. More precisely:
For $i=0,1$, let $K_i$ be knots in $S^3$. A knot concordance from $K_0$ to $K_1$ is a smooth annulus $A=S^1 \times [0,1]$ in $S^3 \times [0,1]$ such that $\partial A= -(K_0) \cup K_1$ where $-$ denotes the reversed orientation.
Using this relation, we can form a group structure on the set of oriented knots in $S^3$, denoted by $\mathcal{C}$.
Let $K$ be a slice knot in $S^3$, that is, $K$ bounds a smooth disk $D$ embedded in $B^4$. We can also show that $K$ is slice if and only if $K$ is concordant to the unknot in $S^3$.
Let $M$ be a closed oriented $3$-manifold. I wonder:
- Can we talk about a knot $M$ behaves like the unknot in $S^3$?
- Can we generalize the notion of knot concordance to the oriented knots in $M$?
- (Extra) Can we define inverses of knots in $M$ as in the case of $S^3$?
Best Answer
One can certainly define concordance by cylinders in $M \times I$ as you suggest. Making the set of concordance classes into a group is problematic, however. Most of the troubles come from the observation that for two knots to be concordant, they must be freely homotopic. This was explored in the Indiana U. PhD thesis of Prudence Heck, Knot concordance in non-simply connected manifolds. I don't think you get a group in any obvious way, for much the same reason that you don't get a group out of the set of free homotopy classes of loops.
There is a notion of homology concordance of oriented pairs $(M,K)$ where $M$ is a homology sphere. The equivalence relation is then concordance in a homology cobordism between $M_0$ and $M_1$. This becomes a group under pairwise connected sum. The unknot in $S^3$ is the 0 element, and pairwise orientation reversal provides inverses.